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3 years ago

PLEASE HELP!!!!!!! What is the mean absolute deviation (MAD) of the data set?

Mathematics

1 answer:

melisa1 [442]3 years ago

7 0

Answer: 2.8

Step-by-step explanation:

The mean of the data set it 10. Then find the difference between the mean and each data value. |10-5|=5 do this for all of them. You get 5,1,1,1,6. Now find the mean of the these new data value. 5+1+1+1+6=14 14/5=2.8

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Do the question in the picture and then tell how you got your answer

liraira [26]

$2(2\sqrt4)^{-2}=\\
2\cdot(2\cdot2)^{-2}=\\
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7 0

3 years ago

What is the next step you should take? 1 + 18n = 33

Ivahew [28]

Answer:

You need to carry the one over by subtracting 1 from 33. The new equation will be 18n = 32

3 0

3 years ago

Read 2 more answers

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago

Determine whether the following relation is a function. {(-6, – 8), (5, 1). (9.-4), (7,2), (15, 0)}

lidiya [134]

This relation is a function. None of the x-values repeat. Each y-value belongs to a specific x-value.

4 0

3 years ago

URGENT! (Grades close tomorrow and turning this in will boost my grade from a B to an A in College Algebra. I can't figure these

zvonat [6]

Answer:

24. see below for a drawing with axis intercepts labeled

30. the plane is parallel to the y-z plane. It is "x=4". The only axis intercept is (4, 0, 0).

Step-by-step explanation:

24. The axis intercepts are fairly easily found. You can do it a couple of ways.

first way

Set other variables to zero and divide by the coefficient of the variable of interest.

For 3x +5y +15z = 15, the x-intercept can be found by setting y and z to zero. This gives ...

3x = 15

so, dividing by 3 tells you the x-intercept is ...

x = 15/3 = 5

Likewise, the y- and z-intercepts are, respectively, ...

y = 15/5 = 3

z = 15/15 = 1

So, the points where the plane crosses the axes are ...

(5, 0, 0), (0, 3, 0), (0, 0, 1) . . . . axes intercepts

second way

Divide by the constant on the right, and express all coefficients as denominators

Doing that gives ...

$\dfrac{3x+5y+15z}{15}=1\\\\\dfrac{x}{5}+\dfrac{y}{3}+\dfrac{z}{1}=1$

The denominators are the corresponding intercepts. (This is called the "intercept form" of the equation of the plane.)

From the above equation, we see the axis intercepts are ...

(5, 0, 0), (0, 3, 0), (0, 0, 1) . . . . same as above

Actually, the math for these two methods is virtually identical. The second way does it "all at once", so can take fewer steps. Effectively, you're doing the math of the "first way" and expressing the result in the denominator.

Please note that when a variable is missing, there is no intercept for that axis. That is, the plane is parallel to that axis. (In problem 30, for example, the y- and z-variables are missing. The plane x=4 is parallel to the y- and z-axes (the y-z plane).)

_____

In the first attached drawing, the x-axis is red, the y-axis is green, and the z-axis is blue. The labeling of the points is difficult to see, but the axes are a different (lighter) color behind the plane than in front of it. This plane forms a triangle in the first octant (x, y, z > 0).

In the second attached drawing, the x=0 plane is blue; the y=0 plane is green, and the z=0 plane is red. The plane of the equation is yellow. This figure better shows the triangle in the first octant.

3-D graph paper (isometric graph paper) grids are available with a web search. Such might make it easier to draw the planes you need to draw. (See the third attachment for one example.)

3 0

3 years ago